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2 years ago

12

The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years. Based on th

e Central Limit Theorem we know the distribution of means from every sample of size 60 will be , with a mean of and a standard deviation of . The probability that a sample mean is 12 or larger for a sample from the horse population is_________

1 answer:

elena-14-01-66 [18.8K]2 years ago

3 0

Answer: 0.9013

Step-by-step explanation:

Given mean, u = 10, standard deviation =8

P(X) =P(Z= X - u /S)

We are to find P(X> or =12)

P(X> or = 12) = P(Z> 12-10/8)

P(Z>=2/8) = P(Z >=0.25)

P(Z) = 1 - P(Z<= 0.25)

We read off Z= 0.25 from the normal distribution table

P(Z) = 1 - 0.0987 = 0.9013

Therefore P(X> or=12) = 0.9013

Note the question was given as an incomplete question the correct and complete question had to be searched online via Google. So the data used are those gotten from the online the Googled question.

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Three highways connect city A with city B. Two highways connect city B with city C.

QveST [7]

(a) The probability that there is no open route from A to B is (0.2)^3 = 0.008.
Therefore the probability that at least one route is open from A to B is given by: 1 - 0.008 = 0.992.
The probability that there is no open route from B to C is (0.2)^2 = 0.04.
Therefore the probability that at least one route is open from B to C is given by:
1 - 0.04 = 0.96.
The probability that at least one route is open from A to C is:
$0.992\times0.96=0.9523$

(b)
α The probability that at least one route is open from A to B would become 0.9984. The probability in (a) will become: $0.9984\times0.96=0.95846$

β The probability that at least one route is open from B to C would become 0.992. The probability in (a) will become:
$0.992\times0.992=0.9841$

Gamma: The probability that a highway between A and C will not be blocked in rush hour is 0.8. We need to find the probability that there is at least one route open from A to C using either a route A to B to C, or the route A to C direct. This is found by using the formula:
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$0.9523+0.8-(0.9523\times0.8)=0.99$
Therefore building a highway direct from A to C gives the highest probability that there is at least one route open from A to C.

4 0

2 years ago

Let p denote the proportion of students at a particular university that use the fitness center on campus on a regular basis. For

9966 [12]

Answer:

a) P-value = 0.0968

b) P-value = 0.2207

c) P-value = 0.0239

d) P-value = 0.0040

e) P-value = 0.5636

Step-by-step explanation:

As the hypothesis are defined with a ">" sign, instead of an "≠", the test is right-tailed.

For this type of test, the P-value is defined as:

$P-value=P(z>z^*)$

being z* the value for each test statistic.

The probability P is calculated from the standard normal distribution.

Then, we can calculate for each case:

(a) 1.30

$P-value=P(z>1.30) = 0.0968$

(b) 0.77

$P-value=P(z>0.77) = 0.2207$

(c) 1.98

$P-value=P(z>1.98) = 0.0239$

(d) 2.65

$P-value=P(z>2.65) = 0.0040$

(e) −0.16

$P-value=P(z>-0.16) = 0.5636$

7 0

2 years ago

Find the quantity q that maximizes profit if the total revenue, R(q), and total cost, C(q) are given in dollars by R(q)=3q−0.002

marysya [2.9K]

Answer:

The value of q that maximize the profit is q=200 units

Step-by-step explanation:

we know that

The profit is equal to the revenue minus the cost

we have

$R(q)=3q-0.002q^{2}$ ---> the revenue

$C(q)=200+2.2q$ ---> the cost

The profit P(q) is equal to

$P(q)=R(q)-C(q)$

substitute the given values

$P(q)=(3q-0.002q^{2})-(200+2.2q)$

$P(q)=3q-0.002q^{2}-200-2.2q$

$P(q)=-0.002q^{2}+0.8q-200$

This is a vertical parabola open downward (because the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the value of q that maximize the profit

The y-coordinate of the vertex represent the maximum profit

using a graphing tool

Graph the quadratic equation

The vertex is the point (200,-120)

see the attached figure

therefore

The value of q that maximize the profit is q=200 units

6 0

2 years ago

What is the finance charge on $13,300 financed at 7.9 percent for 4 years if the monthly payment per $100 is $2.44?

luda_lava [24]

If you borrowed $100, then your monthly payment is $2.44

If you borrowed $200, then your monthly payment is 2*2.44 = 4.88

etc etc

We can set up a proportion

2.44/100 = x/13300

to figure out the monthly payment x. Cross multiply and solve for x

2.44*13300 = 100*x

100x = 2.44*13300

100x = 32452

x = 32452/100

x = 324.52

So the monthly payment is $324.52

An alternative way to get this monthly payment is to apply 2.44% to 13300, which is another way to view the phrase "monthly payment per $100 is 2.44"

------------------

There are 48 months in 4 years (start with 12 mon = 1 yr, then multiply both sides by 4) so we multiply 48 by the monthly payment to get the result 48*324.52 = 15,576.96. This is the total amount you have to pay back which is the principal plus interest.

Subtract off the principal (amount borrowed) to find the interest or finance charge: 15,576.96 - 13,300 = 2,276.96

Answer: Choice B

5 0

2 years ago

Calculate the value of x to one decimal place. inches

kakasveta [241]

Use the law of cosines: $c^2=a^2+b^2-2ab\cos C$

We have:

$c=x\\a=3in\\b=7in\\C=54^o$

substitute:

$\cos54^o\approx0.5878\\\\c^2=3^2+7^2-2\cdot3\cdot7\cdot0.5878\\\\c^2=9+49-24.6876\\\\c^2=33.3129\to c=\sqrt{33.3129}\\\\c\approx5.8$

5 0

2 years ago